Hitung integral tentu berikut

Hitung integral tentu berikut

Jawaban:

$sffrac{2}{3}$

Penjelasan dengan langkah-langkah:

$displaystylesflargeintlimits_{-1}^{0}3{x}^{2}sqrt{{x}^{3}+1}~dx$

$displaystylesflargeint3{x}^{2}sqrt{{x}^{3}+1}~dx$

Misal:

u = x³ + 1

du = 3x² dx

dx = $sffrac{1}{3{x}^{2}}$ du

Penyelesaian:

$displaystylesflargeintcancel{3{x}^{2}}sqrt{u} cdotfrac{1}{cancel{3{x}^{2}}}du$

$displaystylesflargeintsqrt{u}~du$

$displaystylesflargeint{u}^{frac{1}{2}}~du$

$sflarge=frac{1}{frac{1}{2}+1}{u}^{frac{1}{2}+1}$

$sflarge=frac{1}{frac{3}{2}}{u}^{frac{3}{2}}$

$sflarge=frac{2}{3}sqrt{{u}^{3}}$

$sflarge=frac{2}{3}sqrt{{({x}^{3}+1)}^{3}}~Big|_{-1}^{0}$

$sflarge=frac{2}{3}sqrt{{({0}^{3}+1)}^{3}}-(frac{2}{3}sqrt{{({(-1)}^{3}+1)}^{3}})$

$sflarge=frac{2}{3}sqrt{{(0+1)}^{3}}-(frac{2}{3}sqrt{{(-1+1)}^{3}})$

$sflarge=frac{2}{3}sqrt{{(1)}^{3}}-(frac{2}{3}sqrt{{0}^{3}})$

$sflarge=frac{2}{3}sqrt{1}-(frac{2}{3}cdot0)$

$sflarge=frac{2}{3}cdot1-0$

$sflarge=frac{2}{3}$